In triangles ABC and ABD

AB = AB

angle C = angle D

AD = AC. This is sufficient for congruency, (using Pythag. BD = BC so we have three sides (SSS) congruency. )

D is a point on the circumference. C is the given point with ACB = 90

Is it possible to choose D so that AD = AC ?

Well; in both triangles AB is the hypotenuse, and therefore the longest side.

Bisect AB to find the centre of the circle and draw the circumference. With compass point on A and radius AC draw an arc to cut the circle. The arc must cut the circle because AC < AB. Let D be the point where this arc cuts the circle. Then we have found a point D on the circumference with AD = AC.

The theorem states that (A) on any circle with diameter AB, then (B) any point D on the circumference will be such that angle ADB = 90.

So we have A => B But the OP wanted B => A. My congruency proof does this.

Bob

]]>Welcome to the forum.

You'll find the proof that any diameter makes an angle of 90 at the circumference here:

http://www.mathisfunforum.com/viewtopic.php?id=17799

The proof is in post 6.

This is the reverse of what you want but the property in the form A => B also works in reverse, B => A for this theorem.

To prove this just draw the circle and put C somewhere not on the circle along with point D that is. Move D so that AD = AC. ( Think why this is always possible to achieve.) You now have two congruent triangles, ABC and ABD so C and D must coincide.

Bob

]]>(Don't say the angle in a semicircle is a right angle. Need logical proof)

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